Proving that $\lim_{ n \to \infty} \left\{ \frac{1}{2^{m n}} \sum_{r=1}^{2^n-1}(-1)^r r^m\right\}=-\frac{1}{2}$ independently of the value of $m$.

Question

It seems that, independently of the value of $m$, we have

$$ \lim_{ n \to \infty} \left\{ \frac{1}{2^{m n}} \sum_{r=1}^{2^n-1}(-1)^r r^m\right\}=-\frac{1}{2} $$ I've tested it numerically but I have no idea how to prove it.

Can anyone do it? Or give some hints?

Thanks.

I don't know if it would help, but the sum has a formula for given $m\in\Bbb N\cup{0}$. — user170231, Sep 11 '23 at 19:16
By the Euler–Boole summation formula you can show that $$ \sum\limits_{r = 1}^{2^n - 1} {( - 1)^r r^m } = - \frac{1}{2}2^{mn} \left( {1 + \mathcal{O}!\left( {\frac{1}{{2^n }}} \right)} \right). $$ — Gary, Sep 12 '23 at 08:04

Galois group · Accepted Answer · 2023-09-12T07:40:32.173

Let $$\sum\limits_{r=0}^x r^m = P_m(x).$$

Then $$\sum\limits_{r=0}^x (-1)^rr^m = -\sum\limits_{0 \leq r \leq x}r^m+2 \sum\limits_{0 \leq 2r \leq x}(2r)^m=-\sum\limits_{0 \leq r \leq x}r^m+2^{m+1} \sum\limits_{0 \leq r \leq x/2}r^m = -P_m(x)+2^{m+1}P_m([x/2]).$$

It is well-known that $P_m(x)$ is polynomail of degree $m+1$. Let $P_m(x)=c_{m+1}x^{m+1}+c_mx^m+o(x^{m-1})$. Let $x$ be even.(for odd $x$ as in question we can do something similar) Then

$-P_m(x)+2^{m+1}P_m(x/2)=-x^{m+1}c_{m+1}-x^mc_m+2^{m+1}(x/2)^{m+1}c_{m+1}+2^{m+1}(x/2)^{m}c_{m}+o(x^{m-1})=c_m x^{m}+o(x^{m-1}).$

So we only need to know that $c_m = -1/2.$

This is easy if your know properties of bernoulli polynomials. Wiki reads:

$P_m(x)=(B_{m+1}(x+1)-B_{m+1})/(m+1)$. And $$B_{m}(x)=\sum\limits_{k=0}^mC_m^kB_{m-k}x^k.$$

So our statement follows from the formula $B_0=1, B_1=-1/2.$

I tried to answer using the zeta function but it is too tedious. Nice and simple solution (after reading it). — Claude Leibovici, Sep 12 '23 at 06:18

Proving that $\lim_{ n \to \infty} \left\{ \frac{1}{2^{m n}} \sum_{r=1}^{2^n-1}(-1)^r r^m\right\}=-\frac{1}{2}$ independently of the value of $m$.

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